% MAGNETIC VECTOR POTENTIAL
Consider Maxwell's equations (\ref{eqn:geflsfsep3})--(\ref{eqn:alsfsep3}) in a
magnetic source free region, $q_{mv}=\vec{M}_i=0$.
\begin{align}
    \nabla\cdot\varepsilon\vec{E}_A&=q_{ev}\label{eqn:geflA}\\
    \nabla\cdot\mu\vec{H}_A&=0\label{eqn:gmflA}\\
    \nabla\times\vec{E}_A&=-j\omega\mu'\vec{H}_A\label{eqn:flA}\\
    \nabla\times\vec{H}_A&=j\omega\varepsilon'\vec{E}_A+\vec{J}_i\label{eqn:alA}
\end{align} The vector identity
$\nabla\cdot\bigl(\nabla\times\vec A\bigr)=0$, can be used to
define a magnetic vector potential, $\vec A$. Multiplying this
identity by the constant $\mu$ and setting it equal to
(\ref{eqn:gmflA}) yields
$$
    \nabla\cdot\mu\vec{H}_A=
    \nabla\cdot\bigl(\nabla\times\mu\vec{A}\bigr)
    =0
$$
and solving for $\vec H_A$
\begin{equation}\label{eqn:H_A}
    \vec{H}_A
    =\nabla\times\vec{A}
\end{equation}
The subscript A is used as a reminder that the magnetic field was
solved from a magnetic vector potential. The units of $\vec{A}$ is
amperes. Substituting (\ref{eqn:H_A}) into Faraday's law
(\ref{eqn:flA}) yields,
\begin{equation}\label{eqn:1.5sub1.3}
    \nabla\times\vec{E}_A
    =-j\omega\mu'\nabla\times\vec{A}
\end{equation}
%\nabla\times\vec E_A=-\left(j\omega+{\sigma_m\over\mu}\right)\nabla\times\vecA
Using the vector identity, $\nabla\times\bigl(-\nabla\Phi\bigr)=0$
where $\Phi$ is any arbitrary scalar, and rearranging
(\ref{eqn:1.5sub1.3})
$$
    \nabla\times\left[\vec{E}_A+j\omega\mu'\vec{A}\right]
    =\nabla\times\bigl(-\nabla\Phi_e\bigr)=0
$$
Solving for $\vec E_A$ yields an electric field due to an electric
scalar potential, $\Phi_e$ and magnetic vector potential, $\vec
A$.
\begin{equation}\label{eqn:E_A'}
    \vec{E_A}
    =-\nabla\Phi_e-j\omega\mu'\vec{A}
\end{equation}
Taking the curl of both sides of (\ref{eqn:H_A}) and using the
vector identity
$\nabla\times\nabla\times\vec{A}=\nabla\bigl(\nabla\cdot\vec{A}\bigr)-\nabla^2\vec{A}$
\begin{equation}\label{eqn:t1}
    \nabla\times\vec H_A
    =\nabla\times\nabla\times\vec{A}
    =\nabla\bigl(\nabla\cdot\vec A\bigr)-\nabla^2\vec{A}
\end{equation}
Equating (\ref{eqn:t1}) with Ampre's law (\ref{eqn:alA}) yields
\begin{equation}\label{eqn:t2}
j\omega\varepsilon'\vec{E}_A+\vec{J}_i
=\nabla\bigl(\nabla\cdot\vec{A}\bigr)-\nabla^2\vec{A}
\end{equation}
Substituting (\ref{eqn:E_A'}) into (\ref{eqn:t2}) and rearranging
yields the following non homogeneous wave equation
\begin{equation}\label{eqn:waveA0}
\nabla^2\vec{A}+\omega^2\mu'\varepsilon'\vec{A}
=-\vec{J}_i+\nabla\bigl(\nabla\cdot\vec{A}+j\omega\varepsilon'\Phi_e\bigr)
\end{equation}
Since the divergence has not yet been specified, it is chosen
to be
\begin{equation}\label{eqn:lga}
    \nabla\cdot\vec A=-j\omega\varepsilon'\Phi_e
\end{equation}
This is called the Lorentz gauge. $\Phi_e$ is now found to be
\begin{equation}\label{eqn:Phi_e}
    \Phi_e
    =-\frac{\nabla\cdot\vec{A}}{j\omega\varepsilon'}
\end{equation}
Equation (\ref{eqn:waveA0}) now reduces to,
\begin{equation}\label{eqn:waveA}
    \nabla^2\vec{A}-\gamma^2\vec{A}
    =\nabla^2\vec{A}+k^2\vec{A}
    =-\vec{J}_i
\end{equation}
where,
\begin{equation}\label{eqn:gk}
    \gamma^2
    =-k^2
    =-\omega^2\mu'\varepsilon'
\end{equation}
Equation (\ref{eqn:waveA}) is recognized as a non-homogeneous
vector wave equation and (\ref{eqn:gk}) is a complex propagation
constant. Finally, substituting (\ref{eqn:Phi_e}) into
(\ref{eqn:E_A'}) yields the electric fields due to a magnetic
vector potential.
\begin{align}
    \vec{E}_A&=\frac{\nabla\bigl(\nabla\cdot\vec{A}\bigr)}{j\omega\varepsilon'}-j\omega\mu'\vec{A}\label{eqn:E_A1}\\
    &=\frac{1}{j\omega\varepsilon'}\biggl[\nabla\bigl(\nabla\cdot\vec{A}\bigr)-(\gamma^2=-k^2)\vec{A}\biggr]\label{eqn:E_A2}\\
    &=\frac{1}{j\omega\varepsilon'}\biggl[\nabla\bigl(\nabla\cdot\vec{A}\bigr)-\nabla^2\vec{A}-\vec{J}_i\biggr]\label{eqn:E A3}\\
    &=\frac{1}{j\omega\varepsilon'}\biggl[\nabla\times\nabla\times\vec{A}-\vec{J}_i\biggr]\label{eqn:E_A4}
\end{align}
